Problem: $h(n) = -4n+1+5(f(n))$ $f(x) = -3x^{2}-3x$ $ h(f(-2)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = -3(-2)^{2}+(-3)(-2)$ $f(-2) = -6$ Now we know that $f(-2) = -6$ . Let's solve for $h(f(-2))$ , which is $h(-6)$ $h(-6) = (-4)(-6)+1+5(f(-6))$ To solve for the value of $h$ , we need to solve for the value of $f(-6)$ $f(-6) = -3(-6)^{2}+(-3)(-6)$ $f(-6) = -90$ That means $h(-6) = (-4)(-6)+1+(5)(-90)$ $h(-6) = -425$